Dip of the Horizon

Introduction

When we observe sunset and mirage phenomena,
we're usually standing on the surface of the Earth.
But standing on the surface doesn't mean our eyes are at
the surface.
Even at sea, if your eyes were at the surface of the water, your nose would
be under it; this isn't a tenable position for very long.

Diagrams in textbooks are often drawn as if the observer is at the
surface of the Earth — partly because the height of the eye is very small
compared to the size of the Earth: a person's eyes are about 1.6 meters
above the ground, but the radius of the Earth is over 6 million meters.
But, in this case, “very small” isn't the same as “negligible.”
Let's work it out:

Dip, without refraction

Here's a diagram of the situation without atmospheric refraction.
The diagram shows a vertical plane through the center of the Earth (at C)
and the observer (at O).
The radius of the Earth is R, and the observer's eye is a height h
above the point S on the surface.
(Of course, the height of the eye, and consequently the distance to the
horizon, is greatly exaggerated in this diagram.)
The observer's
astronomical horizon
is the dashed line through O, perpendicular to the Earth's radius OC.
Because we are temporarily assuming that there is no refraction, the
observer's
apparent horizon
coincides with the geometric horizon, indicated by
the dashed line OG, tangent to the surface of the Earth.

Because of the observer's height h, the apparent horizon lies below
the astronomical one by the angle dg, which is the geometric
dip of the (unrefracted) horizon.

Because the tangent OG is perpendicular to the radius CG of the circle
that represents the Earth, the angle OCG is also equal to the dip dg,
because both angles labelled dg are complementary to the angle
COG.
So we can apply elementary trigonometry to calculate the dip:

cos (dg) = CG/OC = R/(R + h)

But h is, as remarked above, very small compared to R. That means the
dip angle itself is small; so we can apply the small-angle
approximation cos x = 1 − x2/2
to the left side of the
equation, and use the binomial theorem to expand the right side
[after dividing both numerator and denominator by R, to get
1/(1 + h/R)], which
becomes 1 − h/R.
Then the 1's cancel, leaving us with the approximate
result

(dg)2/2 = h/R ,

or

dg = sqrt (2h/R) .

To get some feeling for how this works, look at some numbers.
Suppose you're standing at the edge of the sea, with your toes at the
waterline. Your eyes are about 1.5 meters above the sea, and
R = 6.4 × 106 m. Then
sqrt(3/6.4 × 106) is about 0.00068 radians, or
0.039°, or about two and a third
minutes of arc.

As the resolution of the normal eye in broad daylight is about one minute
of arc, this dip is an easily visible angle.
So, although the angle is small, it is hardly negligible, even for a
naked-eye observer standing at the water's edge.
Clearly, dip is an appreciable effect for all real-world observations.

Before considering a more realistic situation that includes refraction,
notice that the dip is the same as the angle OCG at the center of the Earth.
A rule of thumb is that a minute of arc on the surface of the Earth is a
(nautical) mile. So this calculation also supplies the
distance to the apparent horizon:
about 2 1/3 nautical miles, in this example.
Equivalently, we can say that as the angular distance to the
horizon is sqrt(2h/R), the linear distance to the horizon is R
times this, or sqrt(2hR).

Dip, including refraction

So far, we have ignored refraction, and drawn the line of sight to the
apparent horizon as the straight line OG. But, in the real world, there
is some terrestrial refraction between the observer and the horizon;
the line of sight to the apparent horizon is not straight, but curved.
So let's try to allow for that curvature:

Usually, the ray is concave toward the Earth, as shown in this drawing.
The solid arc OH now represents the curved line of sight; H is the
(refracted) apparent horizon. Notice that refraction lets us see a little
farther; but the refracted dip d between the horizontal (dashed) and
the tangent to the curved line of sight is now less.
So we no longer have the nice equality between the dip and the distance to
the apparent horizon — although the refraction correction is usually
small, so the geometric relationship is often roughly correct.

The derivation including refraction is a little tedious for a Web page, so
I'll just quote the result here. If we can regard the ray OH as an arc of
a circle, with a curvature k times the Earth's curvature (that is, the
radius of curvature of the ray is R/k), then the above result is still
true if we just replace R in the original expressions (without refraction)
with R/(1 − k).
This is equivalent to saying that the effective curvature of the Earth
is simply diminished by the curvature of the ray: the Earth's curvature is
1/R, the ray's curvature is k/R, and the difference of the curvatures is
(1 − k)/R.

In other words, everything is the same with
refraction as it would be on a fictitious planet with a radius of
R/(1 − k) and no refraction.
So, as long as k isn't very big,
the numbers aren't greatly changed by refraction.

Dip and Temperature

So how big is k? That depends on the temperature gradient; see the
ray-bending
page for details.
It turns out that in “normal” conditions — when the
Standard Atmosphere
is a fair approximation — k is about 1/6 or 1/7 at sea level, and
less on sunny afternoons, or at higher elevations.
Values of k around 0.13 have been used in correcting surveyors' data for a
century or more.
You can find the value of k for different conditions using my JavaScript
calculator.

Clearly, both the dip and the
distance to the horizon are still
proportional to sqrt(h), but the proportionality constant depends on the
value of k adopted — which is equivalent to assuming a particular
lapse rate
between the eye and the surface, as the ray
bending
depends mainly on this temperature gradient.
Typical values used in practice are
dip = 1.75′ × sqrt(h, meters) [taken from the
Explanatory Supplement],
and horizon range = 3.83 km × sqrt(h, meters).

On the other hand, when there is a strong temperature inversion, k can
reach, or exceed, unity. The case k = 1 corresponds to
horizontal rays that orbit the Earth
indefinitely
(a situation considered in exquisite
detail by Kummer.)
Values larger than 1 correspond to ducting conditions; if the observer is
inside the
duct,
a pseudo-horizon appears above the astronomical
one, so the dip of this apparent horizon is negative — a remarkable
phenomenon that really is observed, occasionally.

Caution

Please be aware that the preceding argument contains several unrealistic
assumptions. For example, the lapse rate near the ground is never
constant, so the path of the horizon ray is never an arc of a circle.
Fortunately, in calculating the dip of the horizon, this
idealization makes very little difference. In our
paper
on the dip diagram, George Kattawar and I showed that the dip depends
almost entirely on the difference in temperature between the air at eye
level and that at the apparent horizon, regardless of the thermal
structure in between.

On the other hand, the distance to the horizon turns out to be
sensitive to the intervening thermal structure. That structure tends to
make the distance to the apparent horizon appreciably larger than the
simplistic argument above would suggest.

Furthermore, the air temperature near the
perigee
point on the horizon
ray is considerably closer to the temperature at eye level than is the
temperature of the actual surface. That's because the Earth's surface
isn't smooth: on land, your line of sight is usually intercepted by plants
and structures well above the ground surface; and at sea, the apparent
horizon is at a wave crest, well above the mean surface of the water.
So it's difficult to know what temperature to use for the lowest point
along the horizon ray. (See
here
for more details.)

Even so, it's fair to say that the dip is increased when inferior mirages
are present, and decreased (or even negative) when thermal inversions and
superior mirages occur.

Pictures of variable dip

You can see some examples of dip variations in the green-flash
pictures
taken by Oscar Medina from his home at Pacific Beach.
(The best examples are near the bottom of his Web page.)
Several of his pictures show the roof of a large building nearly level
with the sea horizon. In the ones having an
inferior-mirage flash,
the sea horizon is very nearly even with the top of the building. These
flashes require water warmer than the air.
But those with
mock-mirage flashes,
which are produced by thermal inversions
below eye level, and hence require water considerably colder than the air
at eye level, show a visibly higher sea horizon.

Another
good picture
of thermal effects on dip is shown on
Les Cowley's website.
There, an inferior mirage formed on top of a sunlit sea wall depresses the
apparent horizon well below the normal sea horizon beyond it, making an
illusory depression in the surface of the sea.

These effects are readily visible to the naked eye, and are quite striking
to anyone who regularly observes sunsets from a place that has a
suitable architectural feature nearly at the sea horizon.
They were observed about
1889 by
Annibale Riccò
with the naked eye; but instrumental observations were first made by
Laval (1707),
who found variations of several minutes of arc. Later observations
were made by
Alexander von Humboldt,
who first connected the dip variations with the changes in sea-surface
temperature.

References

You can read more about the dip in papers listed in my
dip file
in the bibliography. I particularly recommend my paper with George
Kattawar on the
dip diagram,
which appeared in Applied Optics37,
3785–3792 (1998).
Several other papers are recommended in my
discussion
of dip variations in connection with calculating the time of sunset;
the works by H. C. Freiesleben, and by Lutz Hasse, on the effects of
waves, are especially important. These effects are briefly mentioned on
another page
as well.